Given that a ＞ 0 and a ≠ 1, x = loga (A3 + 1), y = loga (A2 + 1), try to compare the size of X and y

Given that a ＞ 0 and a ≠ 1, x = loga (A3 + 1), y = loga (A2 + 1), try to compare the size of X and y

The solution ∵ (A3 + 1) - (A2 + 1) = A2 (A-1), (1) when a > 1, A-1 > 0 ∵ A3 + 1 > A2 + 1, because y = logax increases on (0, + ∞), x > y. (2) when 0 < a < 1, A-1 < 0 ∵ A3 + 1 < A2 + 1, because y = logax decreases on (0, + ∞), x > y

Let f (x) = loga (1-x) (a > 0, a ≠ 0)
one
Finding the domain of F (x)
two
Find the value range of X that makes f (x) > 0 hold
The point is that the following steps should be clear····

When 1.1-x > 0, the domain is x1
F (x) = loga (1-x) is an increasing function
So f (x) = loga (1-x) > 0 = loga 1
That is 1-x > 1
x

But when a ＞ 1, given that X1 and X2 are solutions of the equations x + A ^ x = - 1 and X + loga ^ x = - 1 respectively, then X1 + X2 is equal to?
I don't understand,
h(x)=-log(a) x
The sum of the two is the abscissa of the intersection of F (x), G (x) and H (x).
Because g (x) and H (x) are symmetric with respect to y = - x, that is to say, the two functions are inverse functions
So the value of X1 + X2 is equal to the intersection of F (x) = x + 1 and X axis,

X + A ^ x = - 1, that is, x + 1 = - A ^ X
X + log a ^ x = - 1, that is, x + 1 = - log (a) X
Let f (x) = x + 1
g（x）=-a^x
h(x)=-log(a) x
The sum of the two is the abscissa of the intersection of F (x), G (x) and H (x)
Because g (x) and H (x) are symmetric with respect to y = - x, that is to say, the two functions are inverse functions
So the value of X1 + X2 is equal to the intersection of F (x) = x + 1 and X axis,
That is X1 + x2 = - 1

The solution of the equation a (x + m) 2 + B = 0 of X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), then the solution of the equation a (x + m + 2) 2 + B = 0 is______ ．

∵ the solution of the equation a (x + m) 2 + B = 0 about X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), ∵ the equation a (x + m + 2) 2 + B = 0 is transformed into a [(x + 2) + M] 2 + B = 0, that is, in this equation x + 2 = - 2 or x + 2 = 1, the solution is x = - 4 or x = - 1. So the answer is: X3 = - 4, X4 = - 1